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• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering
CHAPTER
6bCHAPMANHALL/CRC
Risk Analysis in Engineering and Economics
Risk Analysis for EngineeringDepartment of Civil and Environmental Engineering
University of Maryland, College Park
ENGINEERING ECONOMICS AND FINANCE
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 1
Economic Equivalence Involving Interest
The Meaning of Equivalence– Economic equivalence is used commonly in
engineering to compare alternatives.– In engineering economy, two things are said to
be equivalent if they have the same effect.– Unlike most individuals involved with personal
finances, corporate and government decision makers using engineering economics might not be so much concerned with the timing of a project's cash flows as with the profitability of the project.
2
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 2
Economic Equivalence Involving Interest
The Meaning of Equivalence (cont’d)– Therefore, analytical tools are needed to
compare projects involving receipts and disbursements occurring at different times, with the goal of identifying an alternative having the largest eventual profitability.
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 3
Economic Equivalence Involving Interest
Equivalence Calculations– Several equivalence calculations are
presented in this section, where these calculations involve the following:1. cash flows,2. interest rates,3. bond prices, and4. loans
3
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 4
Economic Equivalence Involving Interest
Equivalence Calculations (cont’d)– Two cash flows need to be presented along
the same time period using a similar format to facilitate comparison.
– When interest is earned, monetary amounts can be directly added only if they occur at the same point in time.
– Equivalent cash flows are those that have the same value.
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 5
Economic Equivalence Involving Interest
Equivalence Calculations (cont’d)– For loans, the effective interest rate for the
loan, called also the internal rate of return, is defined as the rate that sets the receipts equal to the disbursements on an equivalent basis.
– The equivalence of two cash flows can be assessed at any point in time as illustrated in Example 19.
4
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 6
Economic Equivalence Involving Interest
Example 19: Equivalence Between Cash Flows– Two equivalent cash flows are presented in
Table 5.– The equivalence can be established at any
point in time for an example interest rate of 12% compounded annually.
– For example, if eight years were selected,
for cash flow 1.96.475,2$)12.01(000,1$ 8 =+=F
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 7
Economic Equivalence Involving Interest
Example 19 (cont’d):
$1,573.50$2475.968765
$1,000.00$0.004$0.00$0.003$0.00$0.002$0.00$1,000.001
Cash Flow 2Cash Flow 1YearTable 5. Two Equivalent Cash Flows
5
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 8
Economic Equivalence Involving Interest
Example 19 (cont’d):– While for
cash flow 2.– It should be noted that two or more distinct
cash flows are equivalent if they result into the same amount at the same point in time.
– In this case, the two cash flows are not equivalent.
50.573,1$)12.01(000,1$ 4 =+=F
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 9
Economic Equivalence Involving Interest
Example 20: Internal Rate of Return– According to the equivalence principle, the
actual interest rate earned on an investment can be defined as the interest rate that sets the equivalent receipts to the equivalent disbursements.
– For Table 6, the following equality can be set as:
$1,000+$500(P/F,i,1) + $250(P/F,i,5) = $482(P/A,i,3)(P/F,i,1) + $482(P/A,i,2)(P/F,i,5) (39)
6
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 10
Economic Equivalence Involving Interest
Example 20 (cont’d):
0.00482.0070.00482.006
-250.000.0050.00482.0040.00482.0030.00482.002
-500.000.001-1000.000.000
Disbursements ($)Receipts ($)Time (Year End)Table 6. Converting Cash Flow to its Present Value
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 11
Economic Equivalence Involving Interest
Example 20 (cont’d):– By trial and error i = 10% makes the above
equation valid.– The equivalence can be made at any point of
reference in time; it does not need to be the origin (time = zero) to produce the same answer.
– If the receipts and disbursement of an investment cash flow are equivalent for some interest rate, the cash flows of any two portions of the investment have equal absolute equivalent values at that interest rate.
7
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 12
Economic Equivalence Involving Interest
Example 20 (cont’d):– That is, the negative (-) of the equivalent
amount of one cash flow portion is equal to the equivalent of the remaining portion on the investment.
– breaking up the above cash flow between years 4 and 5, and performing the equivalence at the 4th year produces the following:
-$1,000(F/P,10,4)-$500(F/P,10,3)+$482(F/A,10,3) = -(-$250(P/F,10,1)+$482(P/A,10,2)(P/F,10,1))-$1,000(1.464)-$500(1.331)+$482(3.310) = -(-$250(0.9091)+$482(1.7355)(0.9091))
-$534 = -$534 (40)
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 13
Economic Equivalence Involving Interest
Example 21: Bond Prices– A bond is bought for $900 and has a face
value of $1,000 with 6% annual interest that is paid semiannually.
– The bond matures in 7 years.– The yield to maturity is defined as the rate of
return on the investment for its duration.
8
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 14
Economic Equivalence Involving Interest
Example 21 (cont’d):– Using equivalence, the following equality can
be developed:
– By trial and error i = 3.94% per semiannual period.
– The nominal rate is 2(3.94) = 7.88%, while the effective rate is 8.04%.
$900 = $30(P/A,i,14) + $1,000(P/F,i,14) (41)
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 15
Economic Equivalence Involving Interest
Example 22: Equivalence Calculations for Loans– Suppose a five-year loan of $10,000 (with
interest of 16% compounded quarterly with quarterly payments) is to be paid off after the 13th payment. The quarterly payment is
– The balance can be based on the remaining payments as
$10,000(A/P,4,20) = $10,000(0.0736) = $736
$736(P/A,4,7) = $736(6.0021) = $4,418
(42)
(43)
9
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 16
Economic Equivalence Involving Interest
Amortization Schedule for Loans– An amortization schedule for a loan is defined
as a breakdown of each loan payment (A) into two portions of an interest payment (It) and a payment towards the principal balance (Bt).
– The following terms are defined:
It = interest payment of A at time t,Bt = portion of payment of A to reduce balance at time t
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 17
Economic Equivalence Involving Interest
Amortization Schedule for Loans– The payment can be expressed as
– The balance at end of t-1 is given by
– Therefore, the following relationships can be obtained
A = It + Bt for t = 1, 2, …, n
Bt = A(P/A, i, n-(t-1))
(44)
(45)
Bt = A(P/F, i, n-t+1) (49)
10
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 18
Economic Equivalence Involving Interest
Example 23: Principal and Interest Payments– Suppose a four-year loan of $1,000 (with
interest of 15% compounded annually with annually payments) is to be paid off.
– The payment is A = $1,000(A/P,15,4) = $1,000(0.3503) = $350.265.
– The results are illustrated in Table 7 based on Eq. 49 and using It = A – Bt.
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 19
Economic Equivalence Involving Interest
Example 23 (cont’d):
$401.06$1,000.00$1,401.06Total$45.69$350.265(P/F, 15, 1) = $304.58$350.2654$85.42$350.265(P/F, 15, 2) = $264.85$350.2653
$119.97$350.265(P/F, 15, 3) = $230.30$350.2652$150.00$350.265(P/F, 15, 4) = $200.27$350.2651
Interest Payment (It)
Payment towards Principal (Bt)Loan Payment
($)
Year End
Table 7. Calculations for Example 23
11
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 20
Economic Equivalence Involving Inflation
Price Indexes– For purposes of calculating the effect of
inflation on equivalence, price indexes are used.
– A price index is defined as the ratio between the current price of a commodity or service to the price at some earlier reference time.
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 21
Economic Equivalence Involving Inflation
Example 24: Economic Equivalence Involving Inflation– The base year is 1967, with an index of 100,
and the commodity price is $1.46/lb.– If the price in 1993 is $5.74/lb, the 1993 index
is $5.74/1.4 = $393.20.– The actual consumer price index (CPI) and
annual inflation rates are published and can be used for these computations.
12
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 22
Economic Equivalence Involving Inflation
Annual Inflation Rate– The annual inflation rate at t + 1 can be
computed as:
– The average inflation rate can be computed based on the following condition
t
tttCPI
CPICPI1at rateinflation Annual 1 −=+ + (50)
ntn
t f +=+ CPI)1(CPI (51)
f
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 23
Economic Equivalence Involving Inflation
Annual Inflation Rate (cont’d)– Therefore, the average inflation rate is
(52)1CPI
CPI−= +n
t
ntf
13
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 24
Economic Equivalence Involving Inflation
Example 25: Annual Inflation Rate– Assuming the CPI (of 1966) = 97.2 and the
CPI (of 1980) = 246.80, the average rate of inflation over the 14-year interval can be obtained by applying Eq. 52 as follows:
f = − =
− =
246 8097 2
1246 8097 2
1 6 882%141 14.
...
./
(53)
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 25
Economic Equivalence Involving Inflation
Purchasing Power of Money– The purchasing power at time t in reference to
time period t – n is defined as
– Denoting the annual rate of loss in purchasing power as k, the average rate of loss of purchasing power can be computed as:
(54)t
nttCPI
CPI at timepower Purchasing −=
k
nt
yearbasen
t
yearbase k+
=−CPI
CPI)1(
CPICPI (55)
14
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 26
Economic Equivalence Involving Inflation
Purchasing Power of Money (cont’d)– Solving for CPIt produces the following:
– Therefore,
– Equation 57 relates the average inflation rate and the annual rate of loss in purchasing power .
(56)CPI k CPItn
t n= − +( )1
( )( )
11
1+ =
−f
kn
n
f
k
(57)
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 27
Economic Equivalence Involving Inflation
Constant Dollars– By definition, the constant dollar is
– When using actual dollars, the market interest rate (i) is used.
– When using constant dollars, use the inflation-free interest rate (i*).
Constant Dollars Actual Dollars)=+
11( )
(f n
(58)
15
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 28
Economic Equivalence Involving Inflation
Constant Dollars (cont’d)– The inflation-free interest rate (i*) is defined as
follows for one year:
– For multiple years, it is defined as
(59)111* −++
=fii
( )1
11* −++
= nfii (60)
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 29
Economic Analysis of Alternatives
Present, Annual, and Future-Worth Amounts– The present-worth amount is the difference
between the equivalent receipts and disbursements at the present.
– Assuming Ft to be a net cash flow at time t, the present worth (PW) is
PW i F P F i t F itt
n
tt
t
n( ) ( / , , ) ( )= = +
=
−
=∑ ∑
0 01 (61)
16
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 30
Economic Analysis of Alternatives Present, Annual, and Future-Worth Amounts (cont’d)– The net cash flow Ft is defined as the sum of
all disbursements and receipts at time t.– The annual equivalent amount is the annual
equivalent receipts minus the annual equivalent disbursements of a cash flow.
– It is used for repeated cash flows per year. It is calculated by applying the following equation:
(62)AE i PW i A P i n F ii i
it
t
t
n n
n( ) ( )( / , , ) ( )( )
( )= = +
+
+ −
−
=∑ 1
11 10
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 31
Economic Analysis of Alternatives
Present, Annual, and Future-Worth Amounts (cont’d)– The future worth amount is
– The amounts PW, AE, and FW differ in the point of time used to compare the equivalent amounts.
FW i F F P i n t F itt
n
tn t
t
n( ) ( / , , ) ( )= − = +
=
−
=∑ ∑
0 01 (63)
17
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 32
Economic Analysis of Alternatives
Example 26: Annual Equivalent Amount– The cash flow illustrated in Table 8 is used to
compute the annual equivalent amount based on an interest rate of 10% for a segment of the cash flow that repeats as follows:
– or
AE(10) = [-$1,000+$400(P/F,10,1)+$900(P/F,10,2)](A/P,10,2)
AE(10) = [-$1,000+$400(0.9091)+$900(0.8265)](0.5762) = $61.93
(64)
(65)
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 33
Economic Analysis of Alternatives
Example 26 (cont’d):
MMM
0.00900.00n0.00400.00n-1
-1,000.00900.00n-2
-1,000.00900.0040.00400.003
-1,000.00900.0020.00400.001
-1,000.000.000Disbursements ($)Receipts ($)Year End
Table 8. Cash Flow for Example 26
18
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 34
Economic Analysis of Alternatives
Internal Rate of Return– The internal rate of return (IRR) is the interest
rate that causes the equivalent receipts of a cash flow to be equal to the equivalent disbursements of the cash flow.
– Solving for i* such that the following condition is satisfied:
0 10
= = + −
=∑PW i F it
t
t
n( *) ( *) (66)
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 35
Economic Analysis of Alternatives
Internal Rate of Return (cont’d)– It represents the rate of return on the
unrecovered balance of an investment (or loan).
– The following equation can be developed for loans:
– where U0 is the initial amount of loan or first cost of an asset (F0), Ft is the amount received at the end of the period t, and i* is IRR.
(67)U U i Ft t t= + +−1 1( *)
19
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 36
Economic Analysis of Alternatives
Example 27: Internal Rate of Return– The cash flow illustrated in Table 9 is used to
solve for i by trial and error using the net cash flow and Eq. 66.
– The internal rate of return was determined to be i* = 12.8%.
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 37
Economic Analysis of Alternatives
Example 27 (cont’d)
0.001200.0050.00500.0040.00500.0030.00500.002
-800.000.001-1000.000.000
Disbursements ($)Receipts ($)Year EndTable 9. Cash Flow for Example 27
20
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 38
Economic Analysis of Alternatives
Payback Period– The payback period without interest is the
length of time required to recover the first cost of an investment from the cash flow produced by the investment for an interest rate of zero.
– It can be computed as the smallest n that produces:
Ftt
n
=∑ ≥
00 (68)
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 39
Economic Analysis of Alternatives
Payback Period (cont’d)– The payback period with interest is the length
of time required to recover the first cost of an investment from the cash flow produced by the investment for a given interest rate i.
– It can be computer as the smallest n that produces:
(69)F itt
nt
=
−∑ + ≥0
1 0( )
21
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 40
Economic Analysis of Alternatives
Example 28: Payback Period– According to Table 9, the payback period for
only the $1,000.00 disbursement without interest is 3 years.
– The payback period for only the $1,800.00 disbursement without interest is 5 years.
CHAPTER 6b. ENGINEERING ECONOMICS AND FINANCE Slide No. 41
Homework Assignment #6
Problems:6.66.156.316.336.38
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